Question

If $${\cos }^{-1}\frac{x}{2}+{\cos }^{-1}\frac{y}{3}=\theta ,$$  then 9x² − 12xy cos θ + 4y² is equal to

विकल्प

• 36

•  −36 sin² θ

• 36 sin² θ

• 36 cos² θ

Answer 1

(c) 36 sin² θ

We know
$${\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}[xy-\sqrt{1-x^2}\sqrt{1-y^2}]$$
Now,
$${\cos }^{-1}\frac{x}{2}+{\cos }^{-1}\frac{y}{3}=\theta$$
$$\Rightarrow {\cos }^{-1}[\frac{x}{2}\frac{y}{3}-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{3}}]=\theta$$
$$\Rightarrow \frac{x}{2}\frac{y}{3}-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{3}}=\cos \theta$$
$$\Rightarrow xy-\sqrt{4-x^2}\sqrt{9-y^2}=6\cos \theta$$
$$\Rightarrow \sqrt{4-x^2}\sqrt{9-y^2}=xy-6\cos \theta$$
$$\Rightarrow (4-x^2)(9-y^2)=x^2{y}^2+36{\cos }^2\theta -12xy\cos \theta (\text{ Squaring both the sides })$$
$$\Rightarrow 36-4y^2-9x^2+x^2{y}^2=x^2{y}^2+36{\cos }^2\theta -12xy\cos \theta$$
$$\Rightarrow 36-4y^2-9x^2=36{\cos }^2\theta -12xy\cos \theta$$
$$\Rightarrow 9x^2-12xy\cos \theta +4y^2=36-36{\cos }^2\theta$$
$$\Rightarrow 9x^2-12xy\cos \theta +4y^2=36{\sin }^2\theta$$